Let f(x) = (from 1 to x)∫ sin(1+t^2)dt. Find f'(x)?
Let f(x) = (from 1 to x)∫ sin(1+t^2)dt. Find f'(x)? How exactly do you go about doing this?
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Let f(x) = (from 1 to x)∫ sin(1+t^2)dt. Find f'(x)? How exactly do you go about doing this?
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This is an application of the Fundamental Theorem of Calculus.
Essentially you are taking the derivative of an anti-derivative and as the two are sort of inverses of each other, you are left with the inside evaluated at x: sin(1 + x^2).
If the top limit is a function of x, g(x), then you have to use the chain rule and so if it had been (from 1 to g(x)) of that same integral, then
f ' (x) would have been g ' (x) * sin(1 + x^2).
There are certain rules in doing these types of problems...in this case we just replace the t in sin(1+t^2) with x and the answer is simply sin(1+x^2) times the derivative of x which is one...so anyways...
answer : sin(1 +x^2)